Let $\alpha >0$ and $ \varepsilon > 0 $. Then $\sum_{n = 0}^{ + \infty} \frac{1}{(1 + |n^2 + n - \alpha |) (1 + n + n^2)^{\varepsilon}} < + \infty $? Let $\alpha >0$ and $\varepsilon > 0$. Then $\sum_{n = 0}^{ + \infty} \frac{1}{(1 + |n^2 + n - \alpha |) (1 + n + n^2)^{\varepsilon}} < C <  + \infty $ ? 
where $C $ does not depends on $\alpha$ ?
It seems that traditional methods from calculus are not directly applicable. This problem is related to the nonlinear Schrödinger equation on compact manifolds. 
 A: It suffices to start the summation at $n=1.$ Fix $\alpha >0$ and let $q(x) =x^2 + x -\alpha.$ It is enough to consider
$$\tag 1 \sum_{n=1}^{\infty}\frac{1}{(1+|q(n)|)n^\epsilon}.$$
The unique positive root $r$ of $q$ is $r=(-1 +\sqrt {4\alpha +1})/2.$  We have $q(x) <0$ for $0\le x <r$ and $q(x) >0$ for $x>r.$ Let $n_0\in \{0,1,2,\dots \}$ be the unique integer such that $r\in (n_0,n_0+1].$
We first consider the summation over $n\ge n_0+1.$ This equals
$$\tag 2 \sum_{n=n_0+1}^{\infty}\frac{1}{(1+|q(n)|)n^\epsilon} = \sum_{n=n_0+1}^{\infty}\frac{1}{(1+q(n))n^\epsilon}$$ $$ = \sum_{n=n_0+1}^{\infty}\frac{1}{(1+q(n)-q(r))n^\epsilon}.$$
By the MVT, $q(n)-q(r)=q'(c)(n-r)\ge 1\cdot (n-(n_0+1)).$ Thus the sum in $(2)$ is no more than
$$\sum_{n=n_0+1}^{\infty}\frac{1}{(n-n_0)n^\epsilon}.$$
Letting $n=n_0+k,$ we see the last sum equals
$$ \sum_{k=1}^{\infty}\frac{1}{k(n_0+k)^\epsilon}\le  \sum_{k=1}^{\infty}\frac{1}{k\cdot k^\epsilon}<\infty.$$
For the sum over $n\le n_0,$ we can assume $r\ge 1.$ If not, then for all $n\ge 1,$ $q(n)>r$ and the proof above applies.
For $n<r,$
$$1+|q(n)| = 1-q(n)= 1-(q(n)-q(r))$$ $$=1-q'(c)(n-r)=1+q'(c)(r-n)\ge 1+n_0-n.$$
Thus
$$\sum_{n=1}^{n_0} \frac{1}{(1+|q(n)|)n^\epsilon}\le \sum_{n=1}^{n_0} \frac{1}{(1+n_0-n)n^\epsilon}.$$
I'll be brief here. Break the last sum into two sums, the first over $1\le n <n_0/2,$ the second over $n_0/2\le n\le n_0.$ The first sum is on the order of $1/n_0^\epsilon.$ The second sum is on the order of $(\ln n_0)/n_0^\epsilon.$ Both of these expressions are bounded independent of $n_0.$
Putting this all together, we've shown $(1)$ is bounded independent of $r,$ which is the same as saying $(1)$ is bounded independent of $\alpha.$
